Exercise 5.3.7

$A-B=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$, this is positive semidefinite because it has 0 and 1 as eigenvalues.

$A^2-B^2=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 2\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ The determinant is $-1$ which is less than 0, so it’s not positive semidefinite.